Problem: The equation of hyperbola $H$ is $\dfrac{y^2}{25}-\dfrac {(x-5)^{2}}{16} = 1$. What are the asymptotes?
Answer: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac{y^2}{25} = 1 + \dfrac {(x-5)^{2}}{16}$ Multiply both sides of the equation by $25$ $y^2 = { 25 + \dfrac{ (x-5)^{2} \cdot 25 }{16}}$ Take the square root of both sides. $\sqrt{y^2} = \pm \sqrt { 25 + \dfrac{ (x-5)^{2} \cdot 25 }{16}}$ $ y = \pm \sqrt { 25 + \dfrac{ (x-5)^{2} \cdot 25 }{16}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y \approx \pm \sqrt {\dfrac{ (x-5)^{2} \cdot 25 }{16}}$ $y \approx \pm \left(\dfrac{5 \cdot (x - 5)}{4}\right)$ Rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{5}{4}(x - 5)+ 0$